zoukankan      html  css  js  c++  java
  • POJ 1475 Pushing Boxes

    搜索这样的东西仅仅要写之前规划得当还是蛮顺手的。

    mark[x1][y1][x2][y2]表示小人在(x1,y1) 盒子在(x2,y2)这样的状态是否到过。

    剩下的就是优先队列 + bfs 了。另外开一个栈记录前驱以输出路径。

    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <cstdio>
    #include <queue>
    #include <cmath>
    #include <stack>
    #include <map>
    
    #pragma comment(linker, "/STACK:1024000000");
    #define EPS (1e-8)
    #define LL long long
    #define ULL unsigned long long
    #define _LL __int64
    #define _INF 0x3f3f3f3f
    #define Mod 9999991
    
    using namespace std;
    
    char Map[24][24];
    
    int mark[22][22][22][22];
    
    struct N
    {
        int x1,x2,y1,y2;
    };
    
    struct Q
    {
        int ans,x1,y1,x2,y2,pre;
        int ans2,site;
        bool operator < (const Q &a)const{
            return (a.ans2 < ans2 || (a.ans2 == ans2 && a.ans < ans));
        }
    }q[200000];
    
    bool Judge(Q f)
    {
        if(abs(f.x1-f.x2) + abs(f.y1-f.y2) <= 1)
            return true;
        return false;
    }
    
    int jx[] = {-1, 1, 0, 0};
    int jy[] = { 0, 0,-1, 1};
    
    void Out(int site,Q f)
    {
        if(site == -1)
            return ;
    
        Out(q[site].pre,q[site]);
    
        Q t = q[site];
        
        if(t.x1 == f.x1)
        {
            if(t.y1+1 == f.y1)
            {
                if(t.x2 == f.x2 && t.y2+1 == f.y2)
                {
                    printf("E");
                }
                else
                {
                    printf("e");
                }
            }
            else if(t.y1-1 == f.y1)
            {
                if(t.x2 == f.x2 && t.y2-1 == f.y2)
                {
                    printf("W");
                }
                else
                {
                    printf("w");
                }
            }
        }
        else if(t.y1 == f.y1)
        {
            if(t.x1+1 == f.x1)
            {
                if(t.y2 == f.y2 && t.x2+1 == f.x2)
                {
                    printf("S");
                }
                else
                {
                    printf("s");
                }
            }
            else if(t.x1-1 == f.x1)
            {
                if(t.y2 == f.y2 && t.x2-1 == f.x2)
                {
                    printf("N");
                }
                else
                {
                    printf("n");
                }
            }
        }
    }
    
    void bfs(N s,int n,int m)
    {
        memset(mark,-1,sizeof(mark));
    
        mark[s.x1][s.y1][s.x2][s.y2] = 0;
        priority_queue<Q> pq;
    
        Q f,t;
    
        int S = 0,E = 0;
    
        f.ans = 0;
        f.ans2 = 0;
        f.pre = -1;
        f.x1 = s.x1;
        f.x2 = s.x2;
        f.y1 = s.y1;
        f.y2 = s.y2;
        f.site = 0;
        q[E++] = f;
        pq.push(f);
    
        while(pq.empty() == false)
        {
            f = pq.top();
            pq.pop();
    
            if(Map[f.x2][f.y2] == 'T')
            {
                Out(f.pre,f);
                puts("");
                return ;
            }
    
            t.pre = f.site;
            t.ans = f.ans+1;
            for(int i = 0 ;i < 4; ++i)
            {
                t.x1 = f.x1 + jx[i];
                t.y1 = f.y1 + jy[i];
                t.x2 = f.x2;
                t.y2 = f.y2;
    
                if(1 <= t.x1 && t.x1 <= n && 1 <= t.y1 && t.y1 <= m && Map[t.x1][t.y1] != '#' && -1 == mark[t.x1][t.y1][t.x2][t.y2])
                {
                    if(t.x1 != t.x2 || t.y1 != t.y2)
                    {
                        mark[t.x1][t.y1][t.x2][t.y2] = t.ans;
                        t.site = E;
                        t.ans2 = f.ans2;
                        q[E++] = t;
                        pq.push(t);
                    }
                    else
                    {
                        t.x2 = f.x2 + jx[i];
                        t.y2 = f.y2 + jy[i];
    
                         if(1 <= t.x2 && t.x2 <= n && 1 <= t.y2 && t.y2 <= m && Map[t.x2][t.y2] != '#' && -1 == mark[t.x1][t.y1][t.x2][t.y2])
                         {
                             t.ans2 = f.ans2+1;
                             t.site = E;
                             q[E++] = t;
                             pq.push(t);
                             mark[t.x1][t.y1][t.x2][t.y2] = t.ans;
                         }
                    }
                }
            }
        }
        printf("Impossible.
    ");
    }
    
    int main()
    {
        int i,j,n,m;
    
        int icase = 1;
    
        while(scanf("%d %d",&n,&m) && (n||m))
        {
            for(i = 1;i <= n;++i)
                scanf("%s",Map[i]+1);
    
            N s;
    
            for(i = 1;i <= n;++i)
            {
                for(j = 1;j <= m; ++j)
                {
                    if(Map[i][j] == 'S')
                        s.x1 = i,s.y1 = j;
                    else if(Map[i][j] == 'B')
                        s.x2 = i,s.y2 = j;
                }
            }
            printf("Maze #%d
    ",icase++);
            bfs(s,n,m);
            puts("");
        }
        return 0;
    }
    

  • 相关阅读:
    Fiddler 简介
    jQuery 属性操作
    Win7的虚拟Wi-Fi
    接口与内部类
    继承(二)
    J2EE框架(Struts&Hibernate&Spring)的理解
    继承(一)
    对象与类
    控制流程
    数据类型
  • 原文地址:https://www.cnblogs.com/yxysuanfa/p/6721235.html
Copyright © 2011-2022 走看看